Laws Of Motion Question 3
Question 3 - 25 January - Shift 2
Consider a block kept on an inclined plane (inclined at $45^{\circ}$ ) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu)$ is equal to :
(1) 0.33
(2) 0.60
(3) 0.25
(4) 0.50
Show Answer
Answer: (1)
Solution:
Formula: Second Law Of Motion
$ \begin{aligned} & F_1=\frac{mg}{\sqrt{2}}+\mu mg \cos 45^{\circ} \\ & F_1=\frac{mg}{\sqrt{2}}(1+\mu) \\ & \frac{m_8 sin 45^{\circ}}{45^{\circ}} \\ & F_2=mg \sin 45^{\circ}-f=mg \text{ sin } 45^{\circ}-\mu N \\ & =\frac{mg}{\sqrt{2}}(1-\mu) \\ & F_1=2 F_2 \\ & \frac{mg}{\sqrt{2}}(1+\mu)=2 \frac{mg}{\sqrt{2}}(1-\mu) \\ & 1+\mu=2-2 \mu \\ & \mu=1 / 3=0.33 \end{aligned} $
